Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the real function $f$ defined on the real line $\mathbb{R}$ by $f(x)=x^2$. If $b$ is a given positive real number, show that the restriction of $f$ to the closed interval $[0,b]$ is uniformly continuous by starting with an $e\gt 0$ and exhibiting a $d\gt 0$ which satisfies the requirement of the definition

share|cite|improve this question
What did you try? – Fabian Apr 18 '12 at 19:09
This is a special case of the fact that a continuous function on a compact set is uniformly continuous. – user38268 Apr 18 '12 at 22:40
up vote 2 down vote accepted

A better way to prove this fact comes in the form of the following theorem.

Theorem. Let $K \subseteq \mathbb{R}^N$ compact, and $f:K \to \mathbb{R}$ continuous. Then $f$ is uniformly continuous.

Proof. Let $\epsilon > 0$ be given. By continuity of $f$, for every $x_0 \in K$ choose $\delta(x_0)$ such that if $x \in K$ with $\Vert x-x_0 \Vert < \delta(x_0)$ then $\Vert f(x)- f(x_0) \Vert < \epsilon/2$. The family of open balls $\{B(x,\frac{1}{2}\delta(x)) : x \in K \}$ forms an open cover of $K$. By compactness of $K$ we may choose finitely many $x_1,\ldots, x_n \in K$ such that $\{B(x_i,\frac{1}{2}\delta(x_i)) : 1 \leq i \leq n \}$ still covers $K$. Choose $\delta = \frac{1}{2}\min\{\delta(x_i) : 1 \leq i \leq n \}$. Then for any $x,y \in K$ with $\Vert x - y \Vert< \delta$ there is $i$ with $1\leq i \leq n$ so that $x \in B(x_0,\frac{1}{2}\delta(x_0))$ and $\Vert x_i - y \Vert \leq \Vert x_i - x \Vert + \Vert x-y \Vert < \frac{1}{2}\delta(x_i) + \frac{1}{2}\delta(x_i) \leq \delta(x_i)$. Thus we have $x,y \in B(x_i,\delta(x_i))$ and by the triangle inequality we have $\Vert f(x) -f(y) \Vert \leq \Vert f(x)-f(x_i) \Vert + \Vert f(x_i) - f(y) \Vert < \epsilon/2 + \epsilon/2 = \epsilon. ~~\square$

In the case of this problem, we know that $[0,b]$ is compact for all $b>0$ since it is closed and bounded, and that $x \longmapsto x^2$ is continuous on $[0,b]$. Hence by the previous theorem $x \longmapsto x^2$ is uniformly continuous on $[0,b]$.

Note: if this is a homework problem, this is probably not the solution your professor is looking for, but it is an extremely useful theorem that you might want to know about anyway.

share|cite|improve this answer

If a continuous function f satisfies the property $|f(x_1)-f(x_2)|\le M|x_1-x_2|$ is said to be Lipschitz with constant $M$, and $f$ must be uniformly continuous as in this case you will get $\delta=\frac{\epsilon}{M}$ all the time which will allow your continuous function to be uniformly continuous. For your function we have $|f(x_1)-f(x_2)|=|(x_1^2-x_2^2)|=|(x_1-x_2)(x_1+x_2)|\le 2b|x_1-x_2|$. I hope you can do now.

share|cite|improve this answer

Hint: for $|h|\le1$ and $x\in[0,b]$, we have $$|x^2-(x+h)^2|=|-2xh-h^2|=|h| |h+2x|\le |h|\cdot (2b+1).$$

Given $\epsilon>0$, you essentially find the "$d$ that works" for $x=b$. Then this $d$ will work for all $x\in[0,b]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.